※ 引述《semmy214 (黃小六)》之銘言： : https://imgur.com/a/5gdBShZ : 想請教下這題 令T(x,t)=U(x)+V(x,t) 將穩態跟動態分開 先考慮穩態的部分， 原式變成0=4 partial^2 T/partial x^2 因為穩態時間不變，且V(x,t)在穩態時=0 -> T(x,t)=U(x) -> 0=4 partial^2 U/partial x^2 -> d^2 U/dx^2=0 -> U(x)=ax+b -> U'(x)=a 代入兩個邊界條件 U(0)=b=0 U'(1)=a=1 -> U(x)=x -> T(x,t)=x+V(x,t) 考慮新的邊界條件 T(0,t)= 0+V(0,t)=0 -> V(0,t)=0 partial T(1,t)/partial x = 1+partial V(1,t)/partial x=1 ->partial V(1,t)/partial x=0 將兩個邊界條件都化為0 回到原本問題 partial T(x,t)/partial t = partial V(x,t)/partial t partial^2 T(x,t)/partial x^2 = partial^2 V(x,t)/partial x^2 所以 partial V(x,t)/partial t = 4 partial^2 V(x,t)/partial x^2 -> Z(x)R'(t)=4Z''(x)R(t) -> R'(t)/4R(t)=Z''(x)/Z(x)=λ 先算Z(x) Z''(x)-λZ(x)=0 特徵方程式 α^2-λ=0 令λ=-(β^2) 找共軛虛根 α=+-βi -> Z(x)= C1 cos βx+C2 sin βx 代入邊界條件 Z(0)=C1=0 ->Z(x)=C2 sin βx ->Z'(x)=βC2 cos βx 代入邊界條件 Z'(1)=βC2 cos β=0 因為βC2不能等於0 所以cos β=0 -> β= (2n-1)π/2 (n = 1、2、3、4...) -> Zn(x)=C2n sin (2n-1)π/2 x [其實題目只要算到這就好Eigenfunction] ->λ=-[(2n-1)π/2]^2 [Eigenvalue] 再來算R(t): R'(t)-4λR(t)=0 特徵方程式 α-4λ=0 ->α+[(2n-1)π]^2=0 ->α= -[(2n-1)π]^2 Rn(t)= d1n e^-[(2n-1)π]^2 t Vn(x,t)=Zn(x)Rn(t)= An * e^-[(2n-1)π]^2 t * sin (2n-1)π/2 x (An= d1n*C2n) 因為原方程式是線性方程式，利用重疊定理 V(x,t)=sigma [n=1->∞]Vn(x,t) =sigma [n=1->∞] An * e^-[(2n-1)π]^2 t * sin (2n-1)π/2 x 考慮新的初始條件: T(x,0)=2=x+V(x,0) ->V(x,0)= 2-x = sigma [n=1->∞] An * sin (2n-1)π/2 x [這邊視為用傅立葉sin級數展開，所以(2-x)會被視為奇函數] 積分 [-1->1] (2-x)* sin (2n-1)π/2 x dx = An ->An = 2* 積分 [0->1] (2-x)* sin (2n-1)π/2 x dx = 8/[(2n-1)π]^2 * (-1)^n + 8/[(2n-1)π] 所以V(x,t)=sigma [n=1->∞] An * e^-[(2n-1)π]^2 t * sin (2n-1)π/2 x T(x,t)= x+V(x,t) --

※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 203.204.210.81 (臺灣) ※ 文章網址: https://webptt.com/m.aspx?n=bbs/Math/M.1741528171.A.703.html ※ 編輯: wallowes (203.204.210.81 臺灣), 04/20/2025 00:11:05